Statics and Materials solutionmanual_7b

582

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7–47. Draw the shear and moment diagrams for thesimply supported beam.

300 N/m

4 m

300 N � m

A B

07b Ch07b 582-627.indd 58207b Ch07b 582-627.indd 582 6/12/09 8:59:41 AM6/12/09 8:59:41 AM

583



584


*7–48. Draw the shear and moment diagrams for theoverhang beam.

AB

C

4 m 2 m

8 kN/m


585


•7–49. Draw the shear and moment diagrams for thebeam.

5 m 5 m

2 kN/m50 kN � m

AB

C


586


7–50. Draw the shear and moment diagrams for the beam. 250 lb/ ft

150 lb � ft150 lb � ft

A B

20 ft6 m 250 N · m250 N · m

5 kN/m

6 m

6 m

6 m

30 kN

15 kN

15 kN

250 N · m

250 N · m250 N · m

250 N · m

5 kN/m

15 kN

–15

0.0167 m

–0.250–0.250

0.0167 m

V (kN)

15

22.25M (kN · m)

250 N · m5x

�

+ ΣMA = 0; –30 (3) + By (6) = 0

By = 15 kN

+→ ΣFx = 0; Ax = 0

+↑ΣFy = 0; Ay – 30 + 15 = 0

Ay = 15 kN

For 0 � x � 3 m

+→ ΣFx = 0; 15 – 5x – V = 0

V = 5 (3 – x) Ans

�

+ ΣM = 0; –15 (x) + 0.250 + 5x

2

⎛

⎝⎜

⎞

⎠⎟ + M = 0

M = 1

2(30x – 5x2 – 0.5)

M = 15x – 2.5x2 – 0.25 Ans


587


7–51. Draw the shear and moment diagrams for the beam.

A B

3 m

1.5 kN/m


588


*7–52. Draw the shear and moment diagrams for thesimply supported beam.

A B

150 lb/ft

12 ft

300 lb � ft0.5 kN · m

3 kN/m

4 m

The free – body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is

w = 3x

4

⎛

⎝⎜

⎞

⎠⎟ = 0.75x

Referring to Fig. b,

+↑ΣFy = 0; 1.875 – 1

2(0.75x) (x) – V = 0 V = {1.875 – 0.375x2} kN (1)

�

+ ΣM = 0; M + 1

2(0.75x) (x)

x

3

⎛

⎝⎜

⎞

⎠⎟ – 1.875x = 0 M = {1.875x – 0.125x3} kN · m (2)

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively. The location where the shear is equal to zero can be obtained by setting V = 0 in Eq. (1).

0 = 1.875 – 0.375x2 x = 2.236 m

The value of the moment at x = 2.236 m (V = 0) is evaluated using Eq. (2).

M |x = 2.236 m = 1.875 (2.236) – 0.125 (2.236)3 = 2.795 kN · m


589


12

(3) (4) kN

12

(0.75x) (x) kN

83

m43

m

0.5 kN · m

0.75x

By = 4.125 kNAy = 1.875 kN

Ay = 1.875 kN

V (kN)

1.875

2.236 4

x (m)

–4.125

2.795

4

–0.52.236

x (m)

M (kN · m)


590


•7–53. Draw the shear and moment diagrams for the beam.

AB C

9 ft 4.5 ft

30 lb/ ft

180 lb � ft

M (kN · m)

V (kN)

x (m)

x (m)

0.2x

1.4140.2

4.53

–0.7

3 4.5

0.3

0.2 kN 0.7 kN

0.2 kN

0.9 kN2 m

0.1886

3 m 1.5 m

0.6 kN/m

0.3 kN · m

0 � x � 3 m:

+↑ΣFy = 0; 0.2 – 1

2(0.2x) (x) – V = 0

V = 0.2 – 0.1x2 Ans

V = 0 = 0.2 – 0.1x2

x = 1.414 m

�

+ ΣM = 0; M + 1

2(0.2x) (x)

x

3

⎛

⎝⎜

⎞

⎠⎟ – 0.2x = 0

M = 0.2x – 0.03333x2 Ans

Mmax = 0.2 (1.414) – 0.03333 (1.414)3

= 0.1886 kN · m

3 m < x < 4.5 m:

+↑ΣFy = 0; 0.2 – 0.9 + 0.7 – V = 0

V = 0 Ans

�

+ ΣM = 0; –0.2x + 0.9 (x – 2) – 0.7 (x – 3) + M = 0

M = –0.3 kN · m Ans


591


7–54. If the beam will fail when the maximumshear force is or the maximum moment is

Determine the largest intensity ofthe distributed loading it will support.

wMmax = 1200 lb # ft.Vmax = 800 lb,

L = 18 ft,

L

w

AB

7–54. If L = 5.4 m, the beam will fail when the maximum shear force is Vmax = 4 kN, or the maximum moment is Mmax = 1.8 kN · m. Determine the largest intensity w of the distributed loading it will support.

For 0 � x � L

+↑ΣFy = 0; V = wx

L

2

2 Ans

�

+ ΣM = 0; M = –wx

L

3

6

Vmax = –

2

wL

–4 = – (5.4)

2

w

w = 1.481 kN/m

Mmax = –wL2

6;

–1.8 = – (5.4)

6

2w

w = 0.370 kN/m Ans


592



12 ft

A

12 ft

4 kip/ft

M (kN · m)

V (kN)

x (m)

x (m)

12

(80) (8) = 320 kN

12

(20x) (x) = 10x2

12

[(20) (8 – x)] (8 – x)8 –

3x

4 m 4 m

80 kN/m

MA = 1280 kN · m

Ay = 320 kN

4 m 4 m

20x

20(8 – x)

8 – x

160320

4 8

4 8

–1280

–213.3

Support Reactions : From FBD (a),

�

+ ΣMA = 0; MA – 320 (4) = 0 MA = 1280 kN · m

+↑ΣFy = 0; Ay – 320 = 0 Ay = 320 kN

Shear and Moment Functions : For 0 � x < 4 m [FBD (b)],

+↑ΣFy = 0; 320 – 10x2 – V = 0

V = {320 – 10x2} kN Ans

�

+ ΣM = 0; M + (10x2) x

3

⎛

⎝⎜

⎞

⎠⎟ + 1280 – 320x = 0

M = –1280 + 320 –10

33x x

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ kN · m Ans

For 4 m < x � 8 m [FBD (c)],

+↑ΣFy = 0; V – 1

2[(20) (8 – x)] (8 – x) = 0

V = {(10) (8 – x)2} kN Ans

�+ ΣMD = 0; –

1

2[(20) (8 – x)] (8 – x)

8 –

3

x⎛

⎝⎜

⎞

⎠⎟ – M = 0

M = –10

3(8 – )3x

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ kN · m Ans


593


*7–56. Draw the shear and moment diagrams for thecantilevered beam.

300 lb 200 lb/ft

A

6 ft2 m

4 kN/m1.5 kN

The free – body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is

w = 4x

2

⎛

⎝⎜

⎞

⎠⎟ = 2x

Referring to Fig. b,

+↑ΣFy = 0; –1.5 – 1

2(2x) (x) – V = 0 V = (–1.5 – x2) kN (1)

�

+ ΣM = 0; M + 1

2(2x) (x)

x

3

⎛

⎝⎜

⎞

⎠⎟ + 1.5x = 0 M = –1.5 –

3

3

xx⎛

⎝⎜⎜

⎞

⎠⎟⎟ kN · m (2)

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.


594


M (kN · m)

V (kN)

x (m)

x (m)

12

(4) (2) kN

12

(2) x

13

m23

m

1.5 kN

5.5 kN

1.5 kN

4.333 kN · m

–1.5

–5.5

–4.333


595


•7–57. Draw the shear and moment diagrams for theoverhang beam.

4 kN/m

3 m 3 m

AB


596


7–58. Determine the largest intensity of the distributedload that the beam can support if the beam can withstand amaximum shear force of and a maximumbending moment of .Mmax = 600 lb # ft

Vmax = 1200 lb

w0

w0

2w0

6 ft 6 ft

A B

2 m 2 m

maximum shear force of Vmax = 6 kN and a maximum bending moment of Mmax = 1 kN · m.

Since the loading is discontinuous at the midspan, the shear and moment equations must be written for regions 0 � x < 2 m and 2 m < x � 4 m of the beam. The free – body diagram of the beam’s segment sectioned through the arbitrary point within these two regions are shown in Figs. b and c.

Region 0 � x � 2 m, Fig. b

+↑ΣFy = 0; 2.5w0 – w0x – V = 0 V = w0(2.5 – x) (1)

�

+ ΣM = 0; M + w0xx

2

⎛

⎝⎜

⎞

⎠⎟ – 2.5 w0x = 0 M =

w0

2 (5x – x2) (2)

Region 2 m < x � 4 m, Fig. c

+↑ΣFy = 0; 3.5w0 – 2w0(4 – x) + V = 0 V = w0(4.5 – 2x) (3)

�

+ ΣM = 0; 3.5w0(4 – x) – 2w0(4 – x)1

2(4 – )x

⎡

⎣⎢

⎤

⎦⎥ – M = 0 M = w0(–x2 + 4.5x – 2) (4)

The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of the shear at x = 2 m is evaluated using either Eq. (1) or Eq. (3).

V |x = 2 m = w0(2.5 – 2) = 0.5w0

The location at which the shear is equal to zero is obtained by setting V = 0 in Eq. (3).

0 = w0(4.5 – 2x) x = 2.25 m

The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of the moment at x = 2 m is evaluated using either Eqs. (2) or (4).

M |x = 2 m = w0

2 (10 – 22) = 3w0

The value of the moment at x = 2.25 m (where V = 0) is evaluated using Eq. (4).

M |x = 2.25 m = w0[–2.252 + 4.5 (2.25) – 2] = 3.0625w0

By observing the shear and moment diagrams, we notice that Vmax = 3.5w0 and Mmax = 3.0625w0. Thus,

Vmax = 6 = 3.5w0

w0 = 1.714 kN/m

Mmax = 1 = 3.0625w0

w0 = 0.3265 kN/m (control!) Ans.


597


1 m 2 m 1 m

w0 (2) 2w0 (2)

Ay = 2.5 w0 By = 3.5 w0

Ay = 2.5 w0

By = 3.5 w0

–3.5 w0

0.5 w0

3.0625 w03 w0

0 � x < 2 m

2w0 (4 – x)

4 – x

12

(4 – x)

2 m < x � 4 m

2.5 w0

x (m)

x (m)2.252 4

2

2.25 4


598


7–59. Determine the largest intensity of the distributedload that the beam can support if the beam can withstand amaximum bending moment of and amaximum shear force of .Vmax = 80 kN

Mmax = 20 kN # mw0 w0

4.5 m 1.5 m

AB C


599



600


*7–60. Determine the placement a of the roller support Bso that the maximum moment within the span AB isequivalent to the moment at the support B.

L

a

AB

w0


601


•7–61. The compound beam is fix supported at A, pinconnected at B and supported by a roller at C. Draw theshear and moment diagrams for the beam.

A BC

500 lb/ft

6 ft3 ft1 m 2 m

10 kN/m

The support reactions at A and C and the interaction force at pin connection B are indicated on the free – body diagram of members AB and BC of the compound beam shown in Figs. a and b. Since the loading is continuous through the entire beam and the interaction force at the pin connection at B is internal to the beam, the shear and moment equations can be described by a single function. The free – body diagram of the beam’s left segment sectioned through an arbitrary point is shown in Fig. c.

By referring to Fig. c, we have

+↑ΣFy = 0; 20 – 10x – V = 0 V = {20 – 10x} kN (1)

�

+ ΣM = 0; M + 10xx

2

⎛

⎝⎜

⎞

⎠⎟ + 15 – 20x = 0 M = {20x – 5x2 – 15} kN · m (2)

The shear and moment diagram shown in Figs. d and 3 are plotted using Eqs. (1) and (2), respectively. The location at which the shear is equal to zero is obtained by setting V = 0 in Eq. (1).

0 = 20 – 10x x = 2 m

The value of the moment at x = 2 m (V = 0) is computed using Eq. (2).

M |x = 2 m = 20(2) – 5(22) – 15 = 5 kN · m


602


10 (1) kN

0.5 m 0.5 m

10 (2) kN

1 m 1 m

By = 10 kN

By = 10 kNCy = 10 kN

Ay = 20 kN

MA = 15 kN · m

MA = 15 kN · m

M (kN · m)

x (m)

x (m)

10x

V (kN)

Ay = 20 kN

20

1 2

3

–10

–15

1 2 3

5


603


7–62. The frustum of the cone is cantilevered from pointA. If the cone is made from a material having a specificweight of , determine the internal shear force and momentin the cone as a function of x.

g

A

L x

2 r0

r0


604


7–63. Express the internal shear and moment componentsacting in the rod as a function of y, where 0 … y … 4 ft.

y

z

x

y

4 ft 2 ft

4 lb/ft60 N/m

0.6 m1.2 m

0.3 m

60 (1.2 – y)

1.2 – y

60 (0.6) = 36 N1.2 –2

y

acting in the rod as a function of y, where 0 � y � 1.2 m.

Shear and Moment Functions :

ΣFx = 0; Vx = 0 Ans

ΣFz = 0; Vz – 60 (1.2 – y) – 36 = 0

Vz = {108 – 60y} N Ans

ΣMx = 0; Mx – 60 (1.2 – y)1.2 –

2

y⎛

⎝⎜

⎞

⎠⎟ – 36 (1.2 – y) = 0

Mx = {30y2 – 108y + 86.4} N · m Ans

ΣMy = 0; My – 36 (0.3) = 0

My = 10.8 N · m Ans

ΣMz = 0; Mz = 0 Ans


605


*7–64. Determine the normal force, shear force, andmoment in the curved rod as a function of u.

r

w

u


606


•7–65. The shaft is supported by a smooth thrust bearingat A and a smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft. 300 lb

600 lb400 lb

BA

2 ft 2 ft2 ft 2 ft1 m 1 m 1 m 1 m

3 kN2 kN

1.5 kN

3 kN2 kN 1.5 kN

1 m 1 m 1 m 1 m

Ay = 3.375 kN By = 3.125kN

V (kN)

x (m)

x (m)

M (kN · m)

3.375

1.375

–1.625

–3.125

3.375

4.75

3.125

1 2 3 4

1 2

3 4


607


A B

5 kN10 kN

5 kN

2 m2 m2 m2 m

7–66. Draw the shear and moment diagrams for thedouble overhang beam.


608


7–67. Draw the shear and moment diagrams for theoverhang beam.

AB

M = 10 kN � m2 m 2 m 2 m

6 kN18 kN


609


A B

M � 2 kN � m

4 kN

2 m 2 m 2 m



610


•7–69. Draw the shear and moment diagrams for thesimply supported beam.

AB

2 m 2 m 2 m

10 kN 10 kN

15 kN � m


611


7–70. Draw the shear and moment diagrams for the beam.The support at A offers no resistance to vertical load.

P

L––3

L––3

L––3

A B

P


612


7–71. Draw the shear and moment diagrams for the latheshaft if it is subjected to the loads shown.The bearing at A isa journal bearing, and B is a thrust bearing.

200 mm100 mm 50 mm

50 mm50 mm50 mm200 mm

40 N

80 N60 N 100 N

50 N 40 N 50 N

A B


613


*7–72. Draw the shear and moment diagrams for the beam.

6 m

10 kN

3 kN/m

A B

•7–73. Draw the shear and moment diagrams for theshaft. The support at A is a thrust bearing and at B it is ajournal bearing. A B

2 kN/m4 kN

0.8 m0.2 m


614


7–74. Draw the shear and moment diagrams for the beam. 8 kN15 kN/m

20 kN � m

8 kN

1 m 1 m 1 m0.75 m0.25 m

AB C D


615


7–75. The shaft is supported by a smooth thrust bearing atA and a smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft.

500 N

BA

1.5 m 1.5 m

300 N/m


616


*7–76. Draw the shear and moment diagrams for the beam. 10 kN2 kN/m

5 m 3 m 2 m

A B

•7–77. Draw the shear and moment diagrams for theshaft. The support at A is a journal bearing and at B it is athrust bearing.

1 ft 4 ft 1 ft

100 lb/ft

A 300 lb � ft

200 lb

B

0.3 m 1.2 m 0.3 m

1000 N 2000 N/m

500 N · m

500 N · m

M (N · m)

V (m)

–1000

2000 N/m

1.2 m 0.3 m

2033.33 N 1366.67 N

0.3 m

1000 N

1033.33

–1366.67

0.517 m

–500–33.1

–300

0.817

x (m)

x (m)


617


7–78. The beam consists of two segments pin connected atB. Draw the shear and moment diagrams for the beam.

8 ft 4 ft 6 ft

700 lb150 lb/ft

800 lb � ft

A B C

3.5 kN

2 m

1.2 kN · m3 kN/m

1 m 1.5 m

1.2 kN · m

11.35 kN · m

3 (1.5) kN

0.75 m 0.75 m

2 m 1 m

4.95

1.45

3.483 4.5

–3.05

32

2 33.483

–1.2

0.350

–1.45

–11.35

4.5

M (kN · m)

x (m)

x (m)

V (kN)

3.5 kN

Ay = 4.95 kNBy = 1.45 kN Cy = 3.05 kN


618


7–79. Draw the shear and moment diagrams for thecantilever beam.

300 lb 200 lb/ft

A

6 ft

1.5 kN

2 m

4 kN/m

4 kN/m

1.5 kN

2 m

5.667 kN

–5.5

–5.667

–1.5

5.4 kN · m

x (m)

x (m)

V (kN)

M (kN · m)


619



10 kN

10 kN/m

AB

3 m 3 m


620


•7–81. Draw the shear and moment diagrams for thebeam.

A B

2000 lb500 lb/ ft

9 ft 9 ft

7–82. Draw the shear and moment diagrams for the beam. w0

AB

LL

10 kN

3 m 3 m

10 kN/m

3

3 m3 m

x (m)

x (m)

V (kN)

M (kN · m)

10 kN/m

10 kN

22.5 kN

–22.552.5

–7.5

2.532.5

32.5 kN


621



3 m

8 kN/m

8 kN/m

3 m

A


622


40 kN/m

20 kN

150 kN � m

AB

8 m 3 m



623


•7–85. The beam will fail when the maximum momentis or the maximum shear is Determine the largest intensity w of the distributed load thebeam will support.

Vmax = 8 kip.Mmax = 30 kip # ftw

6 ft 6 ftA

B

2 m 2 m

2 m 23

w

23

m

29w

–23w

43m

43w

–23w

73w

–w

2 m

•7–85. The beam will fail when the maximum moment is Mmax = 50 kN · m or the maximum shear is Vmax = 40 kN. Determine the largest intensity w of the distributed load the beam will support.

Vmax = 4

3

w; 40 =

4

3

w⎛

⎝⎜

⎞

⎠⎟

w = 30 kN/m

Mmax = –2

3

w; –50 = –

2

3

w

w = 75 kN/m

Thus, w = 30 kN/m Ans


624


7–86. Draw the shear and moment diagrams for thecompound beam.

5 kN3 kN/m

AB C D

3 m 3 m 1.5 m 1.5 m


625


7–87. Draw the shear and moment diagrams for the shaft.The supports at A and B are journal bearings.

A B

2 kN/m

300 mm450 mm600 mm


626


A

6 ft 10 ft 6 ft

5 kip/ft

B

15 kip � ft15 kip � ft


3 m 1.8 m1.8 m

25 kN · m 25 kN · m

100 kN/m

25 kN · m

25 kN · m

25 kN · m

12

(100) (1.8) = 90 kN 12

(100) (1.8) = 90 kN

12

(100) (1.8) = 90 kN

x (m)

V (kN)

x (m)

M (kN · m)

1.2 m0.6 m 1.5 m 1.5 m 1.2 m 0.6 m

1.2 m0.6 m

100 (3) = 300 kN

90

150

6.64.8

–150

3.3

–90

6.6

–25

–79–79

3.3

4.81.8

–25

1.8

33.5

Ay = 240 kN By = 240 kN

Support Reactions : From FBD (a),

�

+ ΣMA = 0; By (3) + 90 (0.6) + 25 – 300 (1.5) – 90 (3.6) – 25 = 0

By = 240 kN

+↑ΣFy = 0; Ay + 240 – 90 – 300 – 90 = 0

Ay = 240 kN

Shear and Moment Diagrams : The value of the moment at supports A and B can be evaluated using the method of sections [FBD (c)],

�

+ ΣM = 0; M + 90 (0.6) + 25 = 0

M = –79 kN · m


627


•7–89. Determine the tension in each segment of thecable and the cable’s total length. Set .P = 80 lb

P

A

B

C

D

2 ft

3 ft

50 lb

5 ft

4 ft3 ft

0.6 m

1.5 m

0.9 m1.2 m0.9 m

250 N

0.9 m 1.2 m

0.9 m

250 N

250 N

Ax = 210.112 kN

Ay = 314.763 kN

TBD = 390.935 kN

400 N

cable and the cable’s total length. Set P = 400 N.

From FBD (a)

�

+ ΣMA = 0; TBD cos 59.04° (0.9) + TBD sin 59.04° (2.1) – 250 (2.1) – 400 (0.9) = 0

TBD = 390.935 N = 390.9 N Ans

+→ ΣFx = 0; 390.935 cos 59.04° – Ax = 0

Ax = 210.112 N

+↑ΣFy = 0; Ay + 390.935 sin 59.04° – 400 – 250 = 0

Ay = 314.763 N

Joint A :

+→ ΣFx = 0; TAC cos � – 210.112 = 0 (1)

+↑ΣFy = 0; –TAC sin � + 314.763 = 0 (2)

Solving Eqs. (1) and (2) yields :

� = 56.28°

TAC = 378.4 N Ans

Joint D :

+→ ΣFx = 0; 390.935 cos 59.04° – TCD cos � = 0 (3)

+↑ΣFy = 0; 390.935 sin 59.04° – TCD sin � – 250 = 0 (4)

Solving Eqs. (3) and (4) yields :

� = 22.97°

TCD = 218.4 N Ans

Total length of the cable:

lT = 1.5

sin 59.04° +

1.2

cos 22.97° +

0.9

cos 56.28° = 4.674 m Ans


Statics and Materials solutionmanual_7b

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